Question

When a graph is plotted between log K and $ \left( \frac{1}{T} \right), $ the slope of line obtained represents: (K = rate constant, T = temperature)

• A. $ \frac{{{E}_{a}}}{2.303R} $
• B. $ -\frac{{{E}_{a}}}{R} $
• C. $ -\frac{{{E}_{a}}}{2.303R} $
• D. $ \log A $

Answer 1

Answer: (C)

The Arrhenius equation can be written as $ \log \,K=\log A-\frac{E}{2.303RT} $ On comparing this equation with general equation of a straight line, $ y=mx+c $ We get, $ y=\log \,K, $ $ x=\frac{1}{T}, $ $ m=-\frac{E}{2.303R}, $ $ c=\log A $ i.e., if we plot a graph between log K (at Y-axis) and $ \frac{1}{T} $ (at X-axis), then the slope of the line obtained will be equal to $ -\frac{e}{2.303\,R}. $